ELECTROSTATIC PROBLEMSGiven that all electrostatic mediums are governed by the fact that: We start off by saying that the electric field is irrotational, and that from the null vector identity, We express: We therefore get: This is Poisson’s equation:Where is known as the Laplacian operator –the divergence of a gradient of a scalar. The resulting equation in the different coordinate systems become: Let us show this for the cylindrical and spherical coordinates. And similarly: Edmund Li
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In a simple medium where there is no free charge, then Poisson’s equation reduces to Laplace’s equation:
This equation is the governing equation for problems involving a set of conductors maintained at
different potentials. Once V is found, then can be found and then the charge distribution from .An important fact is that solutions to Laplace’s and Poisson’s equations are unique. We shall
neglect to prove this here.
Example
Consider parallel plate capacitors with a separation of d maintained at potentials 0 to . Assuming
negligible fringing determine the potential and surface charge densities on the plates.
We note that a parallel plate capacitor is a situation of a simple medium and thus we use Laplace’s
equation:
Using the Cartesian coordinates then:
Since the y coordinate is the only space variable then:
Now,
Since the electric field opposes the potential rise then:
The method of replacing bounding surfaces by appropriate image charges in lieu of a formal solution of
Poisson’s or Laplace’s equation is called the method of images.
In order to use this method we must have:
An image charge of opposite polarity
The charges are located in the region of the conducting plane
The conducting plane is equipotential and of infinite extent and depth
We do not want to find the field in the region where image charges are located.
Any given charge configuration above an infinite conducting plane is equivalent tothe combination of given charge configuration and its image configuration with
elimination of the conducting plane.
Example
Consider the point charge shown at a distance d above a grounded xz conducting plane. Find the
voltage at any point P(x,y,z) above the grounded plane.
We first remove the grounded plane and mirror the charge Q to –d with charge –Q. Then:
But note that:
The boundary conditions are:
And the symmetry above the axes:
We can see that our solution satisfies these boundary conditions.
Consider a line charge located at a distance d from the axis of a parallel, conducting, circular cylinder of
radius a. They are both infinitely long.
The image line charge must be inside the cylinder to ensure that the cylindrical surface at r=a is
an equipotential surface
Due to symmetry with respect to OP, the image line must like on the line OP.
The method of images requires us to first assume:
We proceed with this assumption and see if the boundary conditions
are met. The electric potential at distance r from a line charge of
density
is determined to be:
Where is some reference point of zero potential. Using this equation, we can determine the potential
at a point on or outside the cylindrical surface by using the line charge and its image:
Note how the reference point disappears in the equation, and as such we don’t care where the locationof is. Although we have already arrived at a solution, we must find values for and r in terms of the
quantities a and d.
Moreover, to ensure that the surface of the cylinder is equipotential,
This requires that the location of be such that we can make similar triangles as shown .
Thus:
Thus:
We can see that the image line charge – can then replace the cylindrical conducting surface, and V and at any point outside the surface can be determined from the line charges of and – . From this, we
can observe that if we have two transmission lines where is not satisfied, we can treat the problem
as an image problem of two infinitely long line charges of opposite polarity.